● To understand the mole and Avogadro's number.

● To understand equations of state.

● To study the kinetic theory of ideal gas.

● To understand heat capacity.

● To learn and apply the first law of thermodynamics.

● To study thermodynamic processes.

● To understand the properties of an ideal gas.

Chapter 15 Thermal Properties of Matter

• Because atoms and molecules are so small, any practically

meaningful amount of a substance contains a huge number of

atoms or molecules. Therefore, it is more convenient to use a

rather huge measuring unit to “count” their numbers.

• The Avogadro’s number, NA = 6.022×1023 molecules/mole,

is such a measuring unit.

• 1 mole of a pure chemical element or compound contains

NA = 6.022×1023 identical atoms or molecules.

• The molar mass (M) is the mass of 1 mole of a pure chemical

element or compound. It is equal to the Avogadro’s number

multiplying the mass of an atom or molecule (m).

𝑀 = 𝑁𝐴𝑚• The total mass of a system containing n moles of a substance:

𝑚𝑡𝑜𝑡𝑎𝑙 = 𝑛𝑀• The total number of particles in n moles of a substance:

𝑁𝑡𝑜𝑡𝑎𝑙 = 𝑛𝑁𝐴

Example:

Carbon-12 ( 612𝐶)

m = 1.99×10−23 g

(a) The molar mass

𝑀 = 𝑁𝐴𝑚 = 12.0 g

(b) The total mass of 1.5

moles of 612𝐶

𝑚𝑡𝑜𝑡𝑎𝑙 = 𝑛𝑀 = 18.0 g

(c) Number of atoms in 1.5

moles of 612𝐶

𝑁𝑡𝑜𝑡𝑎𝑙 = 𝑛𝑁𝐴= 9.03×1023

15.1 The Mole and Avogadro’s Number

15.2 Equation of State

● Imagine that we can work on the device on the right.

●We may take different actions and expect some results:

If we heat it up, temperature (T) rises and the volume (V)

expands; if we compress it, pressure (p) increases; if we

add more gas (n) into the system, pressure (p) increases

and volume expands (V), etc.

● Question: how are these physical quantities (T, p, V, n, etc.)

related to each other for a given system?

● The equation of state: a mathematical equation relates these

physical quantities to each other. These physical quantities

are also known as state variables or state coordinates.

The Ideal-Gas Equation or Ideal-Gas Law

● For most gases, their state variables very closely obey a simple relationship:

𝑝𝑉 = 𝑛𝑅𝑇.

● R = 8.3145 J/(mol•K) is the ideal-gas constant and is a universal constant for all gasses.

● T is measured in Kelvin (K).

● The above relationship is known as the ideal-gas equation, or, the ideal-gas law.

●We may re-write the ideal-gas equation, replacing n by mtotal/M,

𝑝𝑉 =𝑚𝑡𝑜𝑡𝑎𝑙

𝑀𝑅𝑇

●We may write the ideal-gas equation in terms of the density of the gas, ρ = mtotal/V,

ρ =𝑝𝑀

𝑅𝑇

pV Diagram

T4 > T3 > T2 > T1

𝑝𝑉 = 𝑛𝑅𝑇

Ideal Gas

𝑝 = 𝑛𝑅𝑇/𝑉

Non-ideal gas

pT Phase Diagram

Triple point

and

critical point

15.3 Kinetic Theory of an Idea Gas

● The question: how are measurable macroscopic variables related to microscopic

properties of the atoms and molecules?

● The idea gas: We will treat atoms or molecules as point particles undergoing rapid

elastic collisions with each other and the walls of the container in the given volume.

The potential energies due to all the forces are ignored.

● The process is to apply Newton’s laws to establish the relationship between

microscopic and macroscopic quantities.

● The goals are to understand: the pressure of an ideal gas;

the ideal-gas equation;

the temperature of an ideal gas;

internal energy of an ideal gas;

the heat capacity of an ideal gas;

etc.

Kinetic Molecular Theory of an Idea Gas

Pressure (the impulse of molecule collision with container wall)

(a) momentum change in one collision event: 2𝑚|𝑣𝑥|(b) total momentum change in time interval ∆𝑡:

∆𝑃𝑥 =1

2

𝑁

𝑉𝐴 𝑣𝑥 ∆𝑡 2𝑚 𝑣𝑥 =

𝑁𝐴𝑚𝑣𝑥2∆𝑡

𝑉

(c) Force on the wall is: 𝐹𝑥 =∆𝑃𝑥

∆𝑡=

𝑁𝐴𝑚𝑣𝑥2

𝑉

(d) Pressure on the wall is: 𝑝 =𝐹𝑥

𝐴=

𝑁𝑚𝑣𝑥2

𝑉……(15.6)

(e) Molecules with a distribution of velocities? Take average.

𝑝 =𝐹𝑥

𝐴=

𝑁𝑚(𝑣𝑥2)𝑎𝑣

𝑉=

𝑁

𝑉

1

3𝑚(𝑣2)𝑎𝑣 =

1

𝑉

2

3𝐾𝑡𝑟 ,

where 𝐾𝑡𝑟 = 𝑁[1

2𝑚(𝑣2)𝑎𝑣] is the total kinetic energy.

(f) Compare with the ideal-gas equation: 𝑝𝑉 = 𝑛𝑅𝑇, we have

the total kinetic energy of the gas molecules,

𝐾𝑡𝑟 =3

2𝑛𝑅𝑇 ………………………………(15.7)

The Boltzmann Constant

● The total kinetic energy of all the particles in an ideal gas is 𝐾𝑡𝑟 =3

2𝑛𝑅𝑇.

● It relates the microscopic properties to measurable macroscopic quantities.

● The kinetic is independent of the mass of the atoms or molecules.

● The average kinetic energy per atom or molecule is

𝐾𝑎𝑣 =1

2𝑚(𝑣2)𝑎𝑣=

𝐾𝑡𝑟

𝑛𝑁𝐴=

3

2𝑛𝑅𝑇

𝑛𝑁𝐴=

3

2(𝑅

𝑁𝐴)𝑇.

● Define the Boltzmann constant

𝑘 =𝑅

𝑁𝐴=

8.314 𝐽/(𝑚𝑜𝑙∙𝐾)

6.022×1023/𝑚𝑜𝑙𝑒= 1.381 × 10−23 𝐽/𝐾,

then, 𝐾𝑎𝑣 =1

2𝑚(𝑣2)𝑎𝑣=

3

2𝑘𝑇………….……………..........…(15.8)

which is independent of the details of the particles, such as the mass.

●We can re-write the ideal-gas equation to 𝑝𝑉 = 𝑛𝑅𝑇 = 𝑛 𝑘𝑁𝐴 𝑇,

or 𝑝𝑉 = 𝑁𝑘𝑇 ……………….(15.9)

Maxwell-Boltzmann Distribution

of Molecular Speed

Molecular Speeds in an Ideal Gas

The average kinetic energy per atom or molecule:

𝐾𝑎𝑣 =1

2𝑚(𝑣2)𝑎𝑣 =

3

2𝑘𝑇,

from which we obtain the root-mean-square velocity

𝑣𝑟𝑚𝑠 = (𝑣2)𝑎𝑣 =3𝑘𝑇

𝑚,

where m is the mass of an atom or molecule.

Since 𝑘𝑁𝐴 = 𝑅 and 𝑚𝑁𝐴 = 𝑀, we may re-write,

𝑣𝑟𝑚𝑠 =3𝑅𝑇

𝑀.

Note: (𝑣2)𝑎𝑣 ≠ (𝑣𝑎𝑣)2 = 0

15.4 The Molar Heat Capacities

Note: In Chapter 14, we defined the specific heat as the heat energy required to raise the

temperature of 1 kg of a substance by 1 ºC (or 1 K).

Now, we are going to define a quantity called the molar heat capacity. Its meaning is similar

to that of the specific heat, but it is defined in different units.

Question: How much heat energy 𝑄 is needed to raise n moles of a substance by a

temperature ∆𝑇?

The answer is: 𝑄 = 𝑛𝐶∆𝑇

Note:

(a) The heat energy is proportional to n and ∆𝑇.

(b) The proportionality constant C is called the molar heat capacity. It is the heart energy

needed to raise the temperature of 1 mol of a substance by 1 ºC (or 1 K).

(c) C is in general material-dependent. Yet, it has some simple forms for an ideal gas.

(d) The molar heat capacity has the units of J/(mol•K).

(e) Q is defined positive if it is transferred into the system, and, negative otherwise.

(f) Relationship with the specific heat 𝑐: 𝐶 = 𝑀𝑐 (𝑀 is the molar mass)

Constant-Volume Molar Heat Capacity of an Ideal Gas

Consider an ideal gas with its volume fixed, when heat energy 𝑄 is added to or

remove from the ideal gas, the total kinetic energy 𝐾𝑡𝑟 =3

2𝑛𝑅𝑇 is changed by the

same amount based on the conservation of energy (because, for an ideal gas, the

potential energies due to all the forces are ignored):

𝑄 = ∆𝐾𝑡𝑟 =3

2𝑛𝑅∆𝑇 = 𝑛(

3

2𝑅)∆𝑇.

Therefore, the constant-volume molar heat capacity is

𝐶𝑉 =3

2𝑅 = 12.5 J/(mol•K)

Note:

(a) The constant-volume molar heat capacity 𝐶𝑉 given above is correct for

monatomic gases and is independent of the details of the atoms, such as atomic

masses.

(b) For gases of diatomic molecules, 𝐶𝑉 =5

2𝑅 = 20.8 J/(mol•K)

15.5 The First Law of Thermodynamics

It sets the relationship between the change in the internal

energy ∆𝑈, work done by the system 𝑊, and heat transfer.

𝑄 𝑊∆𝑈

∆𝑈 = 𝑄 −𝑊

Note: (a) 𝑄 is positive if added to the system; negative if removed from the system.

(b) W is positive if it is done by the system on the surrounding; negative otherwise.

Work done during volume change

Work: 𝑊 = 𝐹∆𝑥 = 𝑝𝐴∆𝑥 = 𝑝∆𝑉

Work done at constant pressure: 𝑊 = 𝑝 𝑉2 − 𝑉1

When the pressure is not a constant:

𝑊 = 𝑝1∆𝑉 + 𝑝2∆𝑉+ 𝑝3∆𝑉 +⋯ .

which is the area under the pV diagram

∆𝑉

Work is the area under the pV diagram

Work done at constant temperature

𝑊 = 𝑛𝑅𝑇𝑙𝑛𝑉2𝑉1

Work is the area under the pV diagram.

It depends on the exact path that the

system follows.

Example 15.10 on page 483

Given:

(1) from a to b, 𝑄𝑎𝑏 = 150 J of heat is added to the system

(2) From b to d, 𝑄𝑏𝑑 = 600 J of heat is added to the system

Find: (1) internal energy change from a to b

(2) internal energy change from a to b to d

(3) total heat added to system from a to c to d

Solution:

(1) From a to b 𝑊 = 0 ∆𝑈𝑎𝑏 = 𝑄𝑎𝑏 −𝑊 = 150 J

(1) From b to d, constant pressure,

𝑊𝑏𝑑 = 𝑝 𝑉𝑑 − 𝑉𝑏 = (8.0×104)(5.0×10-3 − 2.0×10-3) = 240 J

∆𝑈𝑎𝑏𝑑 = 𝑄𝑎𝑏𝑑 −𝑊𝑎𝑏𝑑 = (150 + 600) – (0 + 240) = 510 J

(3) 𝑄𝑎𝑐𝑑 = ∆𝑈𝑎𝑐𝑑 +𝑊𝑎𝑐𝑑 = ∆𝑈𝑎𝑏𝑑 +𝑊𝑎𝑐

= 510 + (3.0×104)(5.0×10-3-2.0×10-3) = 510 + 90 = 600 J

The First Law

∆𝑈 = 𝑄 −𝑊

• Adiabatic: no heat transfer in or out of the system

𝑄 = 0

∆𝑈 = −𝑊

• Isochoric: no volume change

𝑉 = constant or ∆𝑉 = 0 or 𝑊 = 0∆𝑈 = 𝑄

• Isobaric: no pressure change.

𝑝 = constant

𝑊 = 𝑝 𝑉2 − 𝑉1• Isothermal: no temperature change.

𝑇 = constant or ∆𝑈 = 0

𝑄 = 𝑊

15.6 Thermodynamics Processes ∆𝑈 = 𝑄 −𝑊

15.7 Properties of an Ideal Gas ∆𝑈 = 𝑄 −𝑊

We learned in Section 15.4 that the constant-volume molar heat capacity for a

monatomic ideal gas is

𝐶𝑉 =3

2𝑅 = 12.5 J/(mol•K)

Question: What is the molar heat capacity for a monatomic ideal gas under a constant

pressure? [ 𝑝 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡; 𝑊 = 𝑝 𝑉2 − 𝑉1 = 𝑝∆𝑉 ]

𝑄 = ∆𝑈 +𝑊 = 𝑛𝐶𝑉∆𝑇 + 𝑝∆𝑉

Since 𝑝𝑉 = 𝑛𝑅𝑇, or 𝑝∆𝑉 = 𝑛𝑅∆𝑇

We have

𝑄 = 𝑛𝐶𝑉∆𝑇 + 𝑝∆𝑉 = 𝑛𝐶𝑉∆𝑇 + 𝑛𝑅∆𝑇 = 𝑛 𝐶𝑉 + 𝑛𝑅 ∆𝑇 = 𝑛𝐶𝑝∆𝑇.

The constant-pressure molar heat capacity is 𝐶𝑝 = 𝐶𝑝 + 𝑅 =5

2𝑅